Find $xyz$ given that $x + z + y = 5$, $x^2 + z^2 + y^2 = 21$, $x^3 + z^3 + y^3 = 80$ I was looking back in my junk, then I found this:

$$x + z + y = 5$$
$$x^2 + z^2 + y^2 = 21$$
$$x^3 + z^3 + y^3 = 80$$
What is the value of $xyz$?
A) $5$
B) $4$
C) $1$
D) $-4$
E) $-5$

It's pretty easy, any chances of solving this question? I already have the
answer for this, but I didn't fully understand.
Thanks for the attention.
 A: x + y + z = 5
On squaring both sides,
$(x + y + z)^2 = 25$
$x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = 25$
$21 + 2xy + 2yz + 2zx = 25$
$2xy + 2yz + 2zx = 25 - 21$
$2xy + 2yz + 2zx = 4$
$xy + yz + zx = 2$
Also,
$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$
Putting values,
$80 - 3xyz = (5)\left[21 - (xy + yz + zx)\right]$
80 - 3xyz = (5)(21 - 2)
80 - 3xyz = 95
-3xyz = 15
xyz = -5
A: Consider the polynomial
$$p(t) = (1-x t)(1-y t)(1-z t)$$
Let's consider the series expansion of $\log\left[p(t)\right]$:
$$\log\left[p(t)\right] =-\sum_{k=1}^{\infty}\frac{S_k}{k} t^k$$
where
$$S_k = x^k + y^k + z^k$$
Since we're given the $S_k$ for $k$ up to $3$ we can write down the series expansion of $\log\left[p(t)\right]$ up to third order in $t$, but that's sufficient to calculate $p(t)$, as it's a third degree polynomial. The coefficient of $t^3$ equals $-xyz$, so we only need to focus on that term. We have:
$$\log\left[p(t)\right] = -\left(5 t +\frac{21}{2} t^2 + \frac{80}{3} t^3+\cdots\right)$$
Exponentiating yields:
$$p(t) = \exp(-5t)\exp\left(-\frac{21}{2}t^2\right)\exp\left(-\frac{80}{3}t^3\right)\times\exp\left[\mathcal{O}(t^4)\right]$$
The $t^3$ term can come in its entirety from the first factor, or we can pick the linear term in $t$ from there and then multiply that by the $t^2$ term from the second factor or we can take the $t^3$ term from the last factor. Adding up the 3 possibilities yields:
$$xyz = \frac{5^3}{3!} -5\times\frac{21}{2} + \frac{80}{3} = -5$$
It is also easy to show that $x^4 + y^4 + z^4 = 333$ by using that the coefficient of the $t^4$ term is zero.
A: We have $$(x+y+z)^3=(x^3+y^3+z^3)+3x(y^2+z^2)+3y(x^2+z^2)+3z(x^2+y^2)+6xyz.$$ Hence
$$125=80+3x(21-x^2)+3y(21-y^2)+3z(21-z^2)+6xyz.$$
This leads to
$$45=63(x+y+z)-3(x^3+y^3+z^3)+6xyz.$$
This gives us $45=315-240+6xyz$, so $6xyz=-30$ and $xyz=-5$.
A: 
\begin{align} x + z + y &= u=5 \tag{1}\label{1} ,\\ x^2 + z^2 + y^2 &= v=21 \tag{2}\label{2},\\ x^3 +z^3 + y^3 &= w=80 \tag{3}\label{3}. \end{align}
What is the value of $xyz$?

Surprisingly, the Ravi substitution
works in this case, despite that
not all the numbers $x,y,z$ are positive,
and hence, the corresponding triangle is "unreal".
So, let
\begin{align} 
a &= y + z
,\quad
b = z + x
,\quad
c = x + y
\tag{4}\label{4}
,\\
x&=\rho-a
,\quad
y=\rho-b
,\quad
z=\rho-c
\tag{5}\label{5}
,
\end{align}
where the triplet $a, b, c$ represents the sides of a triangle
with semiperimeter $\rho$, inradius $r$ and circumradius $R$.
Then
\begin{align} 
x + z + y &= \rho
\tag{6}\label{6}
,\\
x^2 + z^2 + y^2 &= \rho^2-2(r^2+4rR)
\tag{7}\label{7}
,\\
x^3 + z^3 + y^3 &= \rho(\rho^2-12rR)
\tag{8}\label{8}
,\\
xyz&=\rho\,r^2
\tag{9}\label{9}
.
\end{align}
Excluding $rR$ from \eqref{7}-\eqref{8},
we get
\begin{align} 
r^2&=
\tfrac16\,\rho^2-\tfrac12\,v+\tfrac13\,\frac w{\rho}
\tag{10}\label{10}
,
\end{align}
and from \eqref{9} we have the answer
\begin{align} 
xyz&=
\tfrac16\,\rho^3-\tfrac12\,v\rho+\tfrac13\,w
=
\tfrac16\,5^3-\tfrac12\,21\cdot5+\tfrac13\,80
=-5
\tag{11}\label{11}
.
\end{align}
As a bonus, we can find that
\begin{align} 
rR &= \tfrac1{12}\,\frac{\rho^3-w}{\rho}
\tag{12}\label{12}
\end{align}
and $x=\rho-a,\ y=\rho-b,\ z=\rho-c$ are the roots of cubic equation
\begin{align}
x^3-\rho\,x^2+(r^2+4rR)\,x-\rho r^2&=0
\tag{13}\label{13}
,\\
\text{or }\quad
x^3-\rho\,x^2+\tfrac12\,(\rho^2-v)\,x-\tfrac13\,w-\tfrac16\,\rho\,(\rho^2-3v)&=0
\tag{14}\label{14}
,\\
x^3-5\,x^2+2\,x+5&=0
\tag{15}\label{15}
.
\end{align}
One of the roots of \eqref{15} is
\begin{align}
x &= \tfrac53+
\tfrac23\,\sqrt{19}\,\cos\Big(\tfrac13\,\arctan(\tfrac9{25}\,\sqrt{331})\Big)
\approx  4.253418
\tag{16}\label{16}
,\\
\text{the other two are }\quad
y&\approx -0.773387
,\quad
z\approx  1.519969
\tag{17}\label{17}
.
\end{align}
